Chemical Engineering Chapter 9 When the upper steady state is used as the initial

subject Type Homework Help
subject Pages 14
subject Words 1060
subject Authors H. Scott Fogler

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To 530 530 530 530
E 3.0E+04 3.0E+04 3.0E+04 3.0E+04
k 0.3478362 0.3478362 164.49672 164.49672
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 1
Explicit equations as entered by the user
[1] dHr = -30000
[2] V = 48
[3] Cao = .5
P9-16 (b)
From part (a), we find the concentration and temperatures at the points where G(T) and R(T)
intersect.
P9-16 (c)
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See Polymath program P9-16-c.pol
POLYMATH Report
Calculated values of DEQ variables
Variabl
e
Initial
value
Minimal
value
Maximal
value
1
t
0
0
6.
2
Ca
0.0681
0.04847
0.4290427
9
U
150.
150.
150.
1
0
A
250.
250.
250.
1
1
v0
400.
400.
400.
1
2
Ta
530.
530.
530.
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0
Differential equations
1
d(Ca)/d(t) = rA + Ca0 / tau - Ca / tau
Explicit equations
1
Ca0 = 0.5
2
Fa0 = 200
1
1
E = 30000
1
2
k = Ar * exp(-E / R /
T)
1
3
tau = V / v0
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When the upper steady state is used as the initial conditions, the unsteady-state mole balance
shows that this steady state is actually unstable. The concentration increases and the temperature
drops to the lower stable steady state.
P9-16 (d)
P9-16 (e)
Starting at the lower steady state, if T0 is increased to 550 °R, the lower steady state is no longer
stable, but the upper steady state is
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P9-16 (f) Individualized solution
P9-16 (g)
The following Polymath program gives the linear analysis of the problem
A = 1.175, B = 6.03, J = 1600, τ = .12
See Polymath program P9-16-g.pol
POLYMATH Report
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
t
0
0
6.
6.
2
y
2.
0.0176788
4.159719
0.0176788
3
x
0.2
-0.0225558
0.2
-0.0001258
10
Ar
1.416E+12
1.416E+12
1.416E+12
1.416E+12
11
R
1.987
1.987
1.987
1.987
12
U
150.
150.
150.
150.
13
Area
250.
250.
250.
250.
14
v0
400.
400.
400.
400.
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9-46
21
delH
-3.0E+04
-3.0E+04
-3.0E+04
-3.0E+04
22
Na0
24.
24.
24.
24.
Differential equations
1
d(y)/d(t) = (-J * (1 - A) * x + (B - C) * y)
Explicit equations
1
Cpa = 37.5 * 2
2
Ca0 = 0.5
10
Ta = 530
11
V = 48
12
E = 30000
13
k = Ar * exp(-E / R / T)
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P9-16 (f)
Only the lower steady state plot of x1 and y1 will be shown.
P9-17
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P9-17 (a)
See Polymath program P9-17-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 0.2 0.2
Ca 0.3 1.34E-65 0.3 1.34E-65
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra1
Explicit equations as entered by the user
[1] UA = 0
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P9-17 (b)
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P9-17 (c)
P9-18
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P9-18 (a)
P9-18 (b)
Now it would be best to slowly add A to B in a semibatch reactor.
P9-18 (c)
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P9-19
Plotting the data of t vs. T gives the initial and final temperature of the reaction, T0 = 52.5 ˚C and
Tf =166.8 ˚C.
Recalling equation (8-29) with ΔCp = 0, X = 1 and T = Tf gives:
See Polymath program P9-19.pol
POLYMATH Report
Model: lnTdot = a0 + a1*T_inverse_kelvin
Variable
Value
95% confidence
a0
18.88162
2.425175
( )
!"#$=%" 0
TTCpH fiiRx
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ln 18.88
s
T
T
!
= +
P9-20 No solution will be given
CDP9-C
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CDP9-D
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